Problem: $\overline{AB} = 6\sqrt{2}$ $\overline{AC} = {?}$ $A$ $C$ $B$ $6\sqrt{2}$ $?$ $ \sin( \angle BAC ) = \frac{ \sqrt{2}}{2}, \cos( \angle BAC ) = \frac{ \sqrt{2}}{2}, \tan( \angle BAC ) = 1$
Solution: $\overline{AB}$ is the hypotenuse $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the adjacent side so we can use the cos function (CAH) $ \cos( \angle BAC ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\overline{AC}}{\overline{AB}}= \frac{\overline{AC}}{6\sqrt{2}} $ $ \overline{AC}=6\sqrt{2} \cdot \cos( \angle BAC ) = 6\sqrt{2} \cdot \frac{ \sqrt{2}}{2} = 6$